WAVE
SUPERPOSITION
Challenging MCQ questions by The Physics Cafe
Compiled and selected by The Physics Cafe
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1
(a)
Explain what is meant by the term interference.
[2]
(b)
e
fa
A Kundt’s tube is an experimental acoustical instrument that serves to measure the
speed of sound in different medium.
It comprises of a long horizontal tube, containing a fine powder, which is closed at
one end. A loudspeaker connected to a signal generator is positioned at the other
C
s
end. The device is shown in Fig 4.1 and the signal generator is set to a frequency
of 400 Hz.
From the interference of waves resulting in stationary waves being formed, an
c
is
y
h
interesting pattern can be observed as seen in Fig 4.1.
Loudspeaker
Signal
Generator
P
e
400 Hz
(i)
Fig 4.1
[1]
Explain the formation of the piles of fine powder at the positions shown on
Fig 4.1.
(iii)
Piles of powder
Label the positions of nodes (N) and antinodes (A) on Fig. 4.1
h
T
(ii)
123 cm
[3]
Hence or otherwise, determine the speed of the sound waves in that
medium.
Speed of Sound =
m s-1
[2]
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Ans
(a)(i)
Interference is observation in the variation of the intensity of the resultant
wave when two of more waves meet due to the resultant displacement being
given by the vector sum of the displacement of the individual wave.
(b) (i)
Loudspeaker
N A N A N A
Signal
Generator
123 cm
cm
Piles of powder
400 Hz
(b)(ii)
When the sound wave that are produced by the source is reflected at the
glass boundary with a phase difference of .
The incident and reflected wave overlaps and as they have the speed and
frequency, resulting in the formation of nodes and antinodes of a standing
wave if the length of the tube is given by odd multiples of /4.
The piles of power gather at the nodes as at the nodes no oscillation occurs.
(b)(iii)
Wavelength of stationary wave
=123 X 2/3 = 82 cm
Hence speed of wave
= 400 Hz X 0.82 m = 328 = 330 m
2
(a)
The wave nature of light may be demonstrated using the phenomena of
interference.
Describe an experiment to show how interference may be demonstrated using light.
You may use a diagram to illustrate your answer.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
………………………………………………………………………………………………[3]
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(b)
A microphone, connected to a cathode ray oscilloscope (c.r.o.), is placed at a
distance of 1.2 m in front of a loudspeaker connected to a source of fixed frequency.
The intensity of sound measured at this position is 0.65 W m-2. The time base of the
c.r.o. was set at 0.50 ms cm-1 and the display on the c.r.o. is shown in
Fig.8.1.
(i) Determine the frequency of the sound wave.
P
e
C
s
c
is
y
h
Fig. 8.1
e
fa
1 cm
frequency = ……………………………… Hz [2]
(ii) The speed of sound in air is given as 330 m s-1.
Calculate the wavelength of the sound wave from the loudspeaker.
h
T
wavelength = ……………………………… m [2]
(iii) The same loudspeaker is now placed outdoors. A man stands in front of the
loudspeaker at a distance 5.5 m away. Determine the intensity of the sound
that the man hears at this distance.
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intensity = ……………………………… W m-2 [2]
(c)
8.40 m
L1
loudspeakers
sound detector
4.20 m
L2
Fig. 8.2
Two identical loudspeakers, L1 and L2, connected to the same source as (b), are
placed 4.20 m apart. A sound detector is placed 8.40 m away from L1, as shown in
Fig. 8.2.
(i)
Show that the distance between L2 and the sound detector is 9.39 m.
[1]
(ii) Hence, determine whether the intensity at the sound detector is a high or low.
[3]
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e
fa
C
s
(iii) The intensity of sound due to L1 alone at sound detector 8.4 m away is I.
Show, in terms of I, the intensity at the sound detector due to both L1 and L2 is
3.6I.
P
e
c
is
y
h
[3]
(iv) The loudspeaker L2 is now moved towards the sound detector along the axis
h
T
shown in Fig. 8.3.
8.4 m
L1
sound detector
4.2 m
L2
Fig. 8.3
Describe and explain the observation at the sound detector as L2 is moved
towards the sound detector, until L2 is also 8.4 m away from the detector.
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……………………………………………………………………………………………
……………………………………………………………………………………………
……………………………………………………………………………………………
…………………………………………………………………………………………[2]
(d)
Noise cancellation headphones comes with embedded microphones, and actively
makes use of interference to reduce noise that the listener hears.
Explain how this is achieved.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
………………………………………………………………………………………………[2]
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Ans
a
Diagram (if drawn) must show light source (e.g. light / laser), object causing
interference (i.e. double slit) and means of observation (e.g. screen)
Light source must be monochromatic and coherent
OR
Description of setup must mention the above items.
Instead of two bright fringes of the screen, a fringe pattern containing alternating
light and dark fringes is observed.
e
fa
Note: drawing of wavefronts / ripples is not accepted as light wave are in multiple
planes
bi
From Fig. 8.1, period, T = 6 x 0.50 x 10-3 s = 3.0 x 10-3 s
Hence, frequency = 1/T = 333.33 = 333 Hz
bii
Using v = f λ,
330 = 333.33 λ
λ = 0.99 m
biii
Since Intensity  (1 / r)2,
ூ
௥
ଶ
Using ratio, ூఱ.ఱ = ቀ௥భ.మቁ
భ.మ
ଵ.ଶ ଶ
ci
cii
ఱ.ఱ
C
s
c
is
y
h
 ‫ܫ‬ହ.ହ = ቀହ.ହቁ 0.65 = 0.031 ܹ ݉
ିଶ
Using Pythagoras’ Theorem,
Distance = 8.40 2 + 4.20 2 = 9.391
= 9.39 m [shown]
path difference = 9.391 – 8.40
= 0.99 m
P
e
Hence path difference = λ or phase difference = 2π rad
The waves from both speakers will arrive in phase at the sound detector.
h
T
Hence constructive interference occurs and the intensity at the sound detector is a
high.
ciii
Using intensity  1/r2, intensity due to L2 only at 9.39 m away from sound detector
is
‫ܫ‬ଶ
8.4 ଶ
= ൬
൰
‫ܫ‬
9.39
 I2 = 0.8003 I
Since intensity  amplitude2 and letting amplitude of sound due to L1 alone at
detector be a,
ூ
ூమ
௔ ଶ
ூ
௔ ଶ
= ቀ௔ ቁ  ଴.଼଴଴ଷ ூ = ቀ௔ ቁ
మ
 a2 = 0.8946 a
మ
Since from 8cii, constructive interference takes place,
resultant amplitude at sound detector = a + 0.8946a = 1.8946a
Hence,
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‫ܫ‬௥௘௦௨௟௧௔௡௧
1.8946 ܽ ଶ
= ൬
൰
‫ܫ‬
ܽ
 Iresultant = 3.6 I [shown]
civ
Since the path difference is equal to 0 λ when L2 is 8.4 m from sound detector,
there must be a position in between where the path difference is 0.5 λ.
Hence a low intensity (at path difference 0.5 λ) followed by a high intensity (at path
difference 0 λ) will be observed at the sound detector.
OR
As L2 moves to 8.4 m away from sound detector, path difference drops from 1λ to
0.5λ to 0λ.
Hence, intensity of sound drops from loud (maximum) to soft (minimum) to loud
(maximum) respectively.
d
The microphones receive sound from the surrounding and the headphones then
produces the same sound but in antiphase (π radian out of phase).
This causes destructive interference of the sound from the surrounding and
reduces the noise perceived by the listener.
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3
(a) Explain what is meant by a progressive wave.
[1]
e
fa
(b) Sound is propagated in air as a longitudinal progressive wave, in which there is a
repeated sequence of displacements of the air particles. Fig. 2.1a illustrates nine
particles, equally spaced along the line AB, in still air.
1
2
3
4
5
6
7
C
s
A
P
c
is
y
h
Fig. 2.1a
d
P
(i)
P
e
8
9
B
Q
Q
Fig. 2.1b
A sound wave of wavelength equal to the distance between A and B is sent
through the air in the direction of PQ. On line PQ on Fig. 2.1a, draw the possible
positions of the nine particles in the wave relative to their undisturbed positions
h
T
which they occupy in still air, when the sound wave propagates through the
particles.
[1]
(ii) Using (b)(i), sketch a graph showing how the displacement d of the particles
from their undisturbed positions vary along PQ on Fig. 2.1b. Take direction to
the right as positive.
[1]
(iii) A sound wave can also be described in terms of a repeated sequence of
changes in pressure. On Fig. 2.1b, identify, and label with H, a point where the
pressure is the highest. Justify your answer.
[2]
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(iv) A loudspeaker, generating a sound wave with a wavelength magnitude equal to
AB, and a screen are placed at A and B respectively. Describe and explain the
changes, if any, to Fig. 2.1b over time as compared to the current scenario.
[2]
(c) A stereo system in a large hall has two identical speakers, S1 and S2, 1.2 m apart.
The amplitude of the output of each speaker is proportional to the voltage across its
terminals. The voltage input to each speaker is adjusted by means of a balance
control. The arrangement is shown in Fig. 2.2.
C
S1
B
y
O
1.2 m
A
balance
control
S2
15 m
Fig. 2.2 (not to scale)
Initially, the speakers are emitting signals of frequency 1000 Hz which are in phase.
The balance control is set such that there is a voltage of 6 V r.m.s. across each
speaker. An observer hears a loud sound of intensity Imax at A. As he moves along
the line AC, 15 m away from the speakers, he observes that the intensity first falls to
zero at point B, a distance y from A. The speed of sound in air is 330 m s-1.
(i)
Determine the distance y.
[2]
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y = ……………. m
(ii) Determine the next higher frequency of the speaker such that point B would
also be a point of zero intensity.
e
fa
C
s
c
is
y
h
frequency = …………….……. Hz [3]
(iii) With the speakers emitting the original signal frequency of 1000 Hz, the balance
control is now adjusted such that the voltages across S1 and S2 are 3.0 V r.m.s.
and 9.0 V r.m.s. respectively. In terms of Imax, determine the new intensity at
point B.
h
T
P
e
intensity = ………………... [3]
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Ans
(a) A progressive wave is a wave in which energy is transferred from one point to
another by means of vibrations or oscillations within the wave.
(b)
1
2
3
4
5
6
7
8
9
A
B
P
Q
Fig. 2.1a
+ve
d
P
Q
Fig. 2.1b
(i) For part (i), positions of particles must conform to 1 wavelength, with clear
differences in displacements of particles.
(ii) For part (ii), wave drawn must correspond to diagram in (i) and displacement of
particles to the right is taken to be positive on the graph.
(iii) Point H, a point of high pressure, occurs at where particle 5 is as this is where
compressions occur due to neighbouring air particles 4 and 6 coming closer
to one another.
Identifying and labelling of H correctly at particle 5.
(iv) As a stationary wave is being formed, the wave will not progress and each
particle will oscillate with different amplitudes.
In the current scenario, the wave will progress towards Q with each particle
moving left and right, reaching the same amplitude over time.
(c) (i) Using x   D where   v
a
f
330
)( 15 )
1000
y
, x is twice that of y (x = 2y) since B is at minimum intensity
( 1.2  2 )
(
y = 2.1 m
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(ii) Since B is a point of minimum intensity, the path difference of S1 and S2 at point B,
S 1B - S 2 B =
 0 .33

 0 .165 m
2
2
For other wavelengths λ, point B will be at zero intensity if the path difference is
odd multiple of

.
2
Therefore,
e
fa
λ
 0.165 m where n  1
2
λ  0.11m
(2n  1)
Using f  v ,

330
f 

C
s
f = 3000 Hz
(iii) Since amplitude ∝ voltage and amplitude2 ∝ intensity,
initially at A, constructive interference occurs as the two waves meet in phase to
give Imax.
Thus amplitude proportional to 6 V + 6 V = 12 V,
hence Imax ∝ (12 V)2 = 144 V2
At B, destructive interference occurs as two waves meet in antiphase,
Thus amplitude proportional to 6 V - 6 V = 0 V giving Imin = 0.
When the speakers are adjusted to 9 V and 3 V,
the resultant amplitude at B is proportional to 9 V – 3 V = 6 V.
Hence Imin ∝ (6 V)2 = 36 V2
Imin= ¼ Imax
h
T
P
e
c
is
y
h
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